Question: Simplify and expand the following expression: $ \dfrac{2n}{3n - 4}-\dfrac{n + 8}{3n - 10} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3n - 4)(3n - 10)$ Multiply the first term by $\dfrac{3n - 10}{3n - 10}$ $ \begin{align*} \dfrac{2n}{3n - 4} \times \dfrac{3n - 10}{3n - 10} & = \dfrac{(2n)(3n - 10)}{(3n - 4)(3n - 10)} \\ & = \dfrac{6n^2 - 20n}{(3n - 4)(3n - 10)}\end{align*} $ Multiply the second term by $\dfrac{3n - 4}{3n - 4}$ $ \begin{align*} \dfrac{n + 8}{3n - 10} \times \dfrac{3n - 4}{3n - 4} & = \dfrac{(n + 8)(3n - 4)}{(3n - 10)(3n - 4)} \\ & = \dfrac{3n^2 + 20n - 32}{(3n - 10)(3n - 4)}\end{align*} $ Now we have: $ = \dfrac{6n^2 - 20n}{(3n - 4)(3n - 10)} - \dfrac{3n^2 + 20n - 32}{(3n - 10)(3n - 4)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{6n^2 - 20n - (3n^2 + 20n - 32)}{(3n - 4)(3n - 10)} $ $ = \dfrac{6n^2 - 20n - 3n^2 - 20n + 32}{(3n - 4)(3n - 10)} $ $ = \dfrac{3n^2 - 40n + 32}{(3n - 4)(3n - 10)}$ Expand the denominator: $ = \dfrac{3n^2 - 40n + 32}{9n^2 - 42n + 40}$